OK, I’m no expert and I will check my notes and books when I get to work tomorrow, but here’s what I remember.

Given an allele frequency p and p + q = 1, Hardy-Weinberg equilibrium will be reached in one generation of random mating if a certain number of conditions are met (allele frequency is constant in the sexes is the only condition I remember off-hand but there are others). Allele frequency will remain constant at this point. H-W equilibrium is easy to calculate. If there are A and a alleles of a gene, and p is the frequency of A alleles in a gene pool (and obviously a = 1 - A), then

freq(AA) = p[sup]2[/sup]

freq(Aa) = 2*p*q

freq(aa) = q[sup]2[/sup]

This stuff is elementary and I’ve had a handle on since high school. The college and graduate level stuff slips away with every year I’ve not used it. That’s what you are asking about.

Obviously with an evolutionary selection conferred, you will deviate from Hardy-Weinberg in each generation. This page gives a pretty good background on selection; the next chapters deal with selection for heterozygotes and against recessives.

The math is pretty straightforward, though. If you are giving allele A an advantage, then at each generation one calculates a new p (signified as p’) based on the advantage. One counts the alleles available after random mating. If you are selecting against the recessive a, for instance if a gives a 2% disadvantage (s=0.02) as per your example, then both AA and Aa have a selection coefficient of 1 and aa has a selection coefficient of (1-s) or 0.98.

One then counts alleles to calculate the q’. Multiplying through by the fitness coefficient (this is copied pretty much directly from the linked page but I’ll try to explain it in my own words):

Original H-W: p[sup]2[/sup] + 2pq + q[sup]2[/sup] = 1

aa has a 2% disadvantage, so it needs to multiplied by 0.98. There are now fewer alleles surviving in the gene pool – 2% of aas have not survived so the total doesn’t add up to 100%.

p[sup]2[/sup] + 2pq + q[sup]2[/sup]*(0.98) = 1-0.02*q[sup]2[/sup]

One gets a new q’ by counting the number of a alleles now in the population. 1 is contributed by all of the heterozygotes (2pq), 2 by all of the aa homozygotes (2*q[sup]2[/sup]**0.98) over the total number of alleles. Divide these by two because each parent only contributes one gamete to the progeny – only 50% of heterozyogtes’ progeny will inherit that a. Remember that 2% of aas were lost, though!*

q’ = (2pq + 2q[sup]2/sup)/(2(1-0.02*q[sup]2[/sup])

That’s one generation. Rinse, lather, repeat. There are some nice simplifications when the allele is lethal (i.e. the selection coefficient is 0), these are listed on the page.

It is all dependent on your initial allelic frequencies and your selection coefficient. Population size shouldn’t matter that much if it is of significant size, greater than around 200 or so. The smaller the population, the more the variance of the allele frequencies will affect the mating. The example given in chapter 3.5 of the above link is for a population of 2. One very quickly sees alleles become lost or fixed. This starts to get beyond my statistical remembrance so I will leave it to you to work through that page.

Hope that’s of some help…